【数学I】例題4.2.2：余弦定理（One More）★★

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% 例題I4.2.2：余弦定理（One More）★★
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$\triangle\mathrm{ABC}$において，次の値を求めよ．

(1) $b=5\sqrt{2},c=8,A=45^{\circ}$のとき，$a$

(2) $a=\sqrt{5},b=\sqrt{2},c=1$のとき，$A$

(3) $a=2\sqrt{6},b=4,A=60^{\circ}$のとき，$c$

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% 例題I4.2.2の解答（One More）★★
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(1) 余弦定理より，

\begin{align*}
a^2&=(5\sqrt{2})^2+8^2-2\cdot 5\sqrt{2}\cdot 8\cdot\cos 45^{\circ}\\
&=50+64-80\sqrt{2}\cdot\frac{1}{\sqrt{2}}\\
&=114-80=34
\end{align*}

よって，$a>0$より，$a=\sqrt{34}$

(2) 余弦定理より，

\begin{align*}
\cos A&=\frac{(\sqrt{2})^2+1^2-(\sqrt{5})^2}{2\cdot\sqrt{2}\cdot 1}\\
&=\frac{2+1-5}{2\sqrt{2}}=\frac{-2}{2\sqrt{2}}=-\frac{1}{\sqrt{2}}
\end{align*}

よって，$0^{\circ}<A<180^{\circ}$であるから，$A=135^{\circ}$

(3) 余弦定理より，$(2\sqrt{6})^2=4^2+c^2-2\cdot 4\cdot c\cdot\cos 60^{\circ}$

したがって，$c^2-4c-8=0$

これを解くと，$c=2\pm 2\sqrt{3}$

よって，$c>0$より，$c=2+2\sqrt{3}$

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% 問題I4.2.2：余弦定理（One More）★★
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$\triangle\mathrm{ABC}$において，次の値を求めよ．

(1) $a=6,b=4,C=120^{\circ}$のとき，$c$

(2) $a=5,b=7,c=8$のとき，$B$

(3) $a=2,c=\sqrt{6},C=60^{\circ}$のとき，$b$

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(1) 余弦定理より，

\begin{align*}
c^2&=6^2+4^2-2\cdot 6\cdot 4\cdot\cos 120^{\circ}\\
&=36+16+24=76
\end{align*}

よって，$c>0$より，$c=2\sqrt{19}$

(2) 余弦定理より，

\begin{align*}
\cos B&=\frac{5^2+8^2-7^2}{2\cdot 5\cdot 8}\\
&=\frac{25+64-49}{80}=\frac{40}{80}=\frac{1}{2}
\end{align*}

よって，$B=60^{\circ}$

(3) 余弦定理より，

\[
(\sqrt{6})^2=2^2+b^2-2\cdot 2\cdot b\cdot\cos 60^{\circ}
\]

したがって，$b^2-2b-2=0$

これを解くと，$b=1\pm\sqrt{3}$

よって，$b>0$より，$b=1+\sqrt{3}$

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% 例題I4.2.2の側注：余弦定理（One More）★★
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% 側注1
$a^2=b^2+c^2-2bc\cos A$

% 側注2
辺の長さは正であるから，$a>0$

% 側注3
$\cos A=\frac{b^2+c^2-a^2}{2bc}$

% 側注4
$a^2=b^2+c^2-2bc\cos A$

% 側注5
辺の長さは正であるから，$c>0$

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% 問題I4.2.2の側注：余弦定理（One More）★★
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% 側注1
$c^2=a^2+b^2-2ab\cos C$

% 側注2
辺の長さは正であるから，$c>0$

% 側注3
$\cos B=\frac{a^2+c^2-b^2}{2ac}$

% 側注4
$c^2=a^2+b^2-2ab\cos C$

% 側注5
辺の長さは正であるから，$b>0$

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% 例題I4.2.2の考え方・補足など：余弦定理（One More）★★
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% 考え方
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余弦定理を用いる．このとき，与えられた辺や角に応じて，(i)の形や(ii)の形の等式を使うとよい．

(1) $a^2=b^2+c^2-2bc\cos A$を用いる．

(2) $\cos A=\frac{b^2+c^2-a^2}{2bc}$を用いる．

(3) $a^2=b^2+c^2-2bc\cos A$を用いる．

[width=5cm]

(i)

\[
a^2=b^2+c^2-2bc\cos A
\]

(ii)

\[
\cos A=\frac{b^2+c^2-a^2}{2bc}
\]

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\[
a^2=b^2+c^2-2bc\cos{A},b^2=c^2+a^2-2ca\cos{B},c^2=a^2+b^2-2ab\cos{C}
\]

\[
\cos{A}=\frac{b^2+c^2-a^2}{2bc},\cos{B}=\frac{c^2+a^2-b^2}{2ca},\cos{C}=\frac{a^2+b^2-c^2}{2ab}
\]